Question 1050303: Find the general equation of the parabola with axis parallel to x-axis and passing through the points (-1,3/2), (1,0) and (-3,-1)
I have reached to the point of :
9/4a + 3/2b + c = -1
C = 1
a - b + c = -3
I don't know what's the next step I used the formula: X = Ay^2 + By + C
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Find the general equation of the parabola with axis parallel to x-axis and passing through the points (-1,3/2), (1,0) and (-3,-1)
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Form:: ay^2 + by + c = x
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Using (-1,3/2) you get: (9/4)a + (3/2)b + c = -1
Using (1,0) you get : 0 + 0 + c = 1
Using (-3,-1) you get a - b + c = -3
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c = 1
So you get:
(9/4)a + (3/2)b = -2
a - b = -4
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Substitute for "a" and solve for "b"::
(9/4)(b-4) + (3/2)b = -2
(9/4)b - 9 + (6/4)b = -2
(3/4)b = 7
b = 28/3
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Solve for "a"::
a = b-4
a = (28/3)-(12/3)
a = 16/3
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Equation::
x = (16/3)y^2 + (28/3)y + 1
OR
x = (1/3)(16y^2 + 28y + 3)
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Cheers,
Stan H.
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